# When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2, how do you determine the remainder when the polynomial is divided by (x+2)(x-1)?

Jul 1, 2015

The remainder when dividing $f \left(x\right)$ by $\left(x + 2\right) \left(x - 1\right)$ is $\left(7 x - 5\right)$

#### Explanation:

Note that $\left(x + 2\right) \left(x - 1\right) = {x}^{2} + x - 2$ is quadratic, so if we divide another polynomial by it then the remainder will be $0$, a non-zero constant or a linear polynomial. Any remainder of greater degree could be divided further.

Suppose $f \left(x\right) = \left(x + 2\right) \left(x - 1\right) k \left(x\right) + r x + s$

Then:

$f \frac{x}{x + 2} = \left(x - 1\right) k \left(x\right) + \frac{r x + s}{x + 2}$

$= \left(x - 1\right) k \left(x\right) + \frac{r x + 2 r - 2 r + s}{x + 2}$

$= \left(x - 1\right) k \left(x\right) + r + \frac{s - 2 r}{x + 2}$

Hence [1]: $s - 2 r = - 19$

$f \frac{x}{x - 1} = \left(x + 2\right) k \left(x\right) + \frac{r x + s}{x - 1}$

$= \left(x + 2\right) k \left(x\right) + \frac{r x - r + r + s}{x - 1}$

$= \left(x + 2\right) k \left(x\right) + r + \frac{r + s}{x - 1}$

Hence [2]: $r + s = 2$

Subtract [2] from [1] to get $- 3 r = - 21$, hence $r = 7$

Then $s = 2 - r = - 5$

So the remainder when dividing $f \left(x\right)$ by $\left(x + 2\right) \left(x - 1\right)$ is $7 x - 5$

Jul 2, 2015

The remainder is $7 x - 5$.

#### Explanation:

Call the polynomial $P \left(x\right)$.

Because $P \left(x\right)$ divided by $x + 2$ leaves a remainder of $- 19$, we know that:
$P \left(x\right) = \left(x + 2\right) {Q}_{1} \left(x\right) - 19$. (Division Algorithm)

Because $P \left(x\right)$ divided by $x - 1$ leaves a remainder of $2$, we know that $P \left(1\right) = 2$ (Remainder Theorem)

From these two facts we get:

$P \left(1\right) = \left(1 + 2\right) {Q}_{1} \left(1\right) - 19 = 2$

So $3 {Q}_{1} \left(1\right) = 21$

and ${Q}_{1} \left(1\right) = 7$

Applying the Division Algorithm and the Remainder Theorem to ${Q}_{1} \left(x\right)$ we learn that:

${Q}_{1} \left(x\right) = \left(x - 1\right) {Q}_{2} \left(x\right) + 7$

Subtitutuing in $P \left(x\right) = \left(x + 2\right) {Q}_{1} \left(x\right) - 19$, we get:

$P \left(x\right) = \left(x + 2\right) \left[\left(x - 1\right) {Q}_{2} \left(x\right) + 7\right] - 19$

$= \left(x + 2\right) \left(x - 1\right) {Q}_{2} \left(x\right) + 7 \left(x + 2\right) - 19$

$= \left(x + 2\right) \left(x - 1\right) {Q}_{2} \left(x\right) + 7 x + 14 - 19$

$= \left(x + 2\right) \left(x - 1\right) {Q}_{2} \left(x\right) + \left[7 x - 5\right]$

The remainder is $7 x - 5$.

May 28, 2017

$7 x - 5.$

#### Explanation:

Let the Poly. in question be $p \left(x\right) .$

We know that, the Degree of the Remainder Poly. is strictly

less than that of the Divisor Poly.

So, when $p \left(x\right)$ is diveded by a Quadr. Poly. $\left(x + 2\right) \left(x - 1\right)$, the

Remainder Ploly. must have a degree $1 , \mathmr{and} , 0.$

Therefore, let us suppose that, the desired remainder is, $a x + b .$

Symbolically, this can be said, as, let,

$p \left(x\right) = \left(x + 2\right) \left(x - 1\right) q \left(x\right) + \left(a x + b\right) \ldots \ldots \ldots \ldots \ldots \ldots \left(\ast\right) .$

Now, $p \left(x\right) ,$ when divided by $\left(x + 2\right)$ leaves remainder $- 19 ,$

$\Rightarrow p \left(- 2\right) = - 19.$

$\Rightarrow - 2 a + b = - 19 , \ldots \ldots \ldots \ldots \left[\because , \left(\ast\right)\right] \ldots \ldots \ldots . . \left({\ast}_{1}\right) .$

On the same lines, $p \left(1\right) = 2 \Rightarrow a + b = 2. \ldots \ldots \ldots \ldots \left({\ast}_{2}\right) .$

Solving $\left({\ast}_{1}\right) \mathmr{and} \left({\ast}_{2}\right) , a = 7 , b = - 5 ,$ giving the desired

remainder, $7 x - 5 ,$ as Respected JIM H. and George C. have